Final answer:
To find the symmetric equations for the line of intersection of the planes 2x + y - z = 3 and 3x - 4y + z = 6, we need to find a direction vector for the line. The parametric equations for the line are x = 3/2 + 5t, y = -3/2 - 7t, and z = -10t. The acute angle between the two planes is approximately 43.77 degrees.
Step-by-step explanation:
To find the symmetric equations for the line of intersection of the planes 2x + y - z = 3 and 3x - 4y + z = 6, we need to find a direction vector for the line.
To do this, we can find the cross product of the normal vectors of the two planes. The normal vector of the first plane is [2, 1, -1] and the normal vector of the second plane is [3, -4, 1]. Taking the cross product of these vectors gives us the direction vector [5, -7, -10].
Next, we can find a point on the line of intersection by setting z to 0 in one of the plane equations and solving for x and y. Let's set z = 0 in the first equation: 2x + y - 0 = 3. Solving this equation gives us x = 3/2 and y = -3/2.
Therefore, the parametric equations for the line of intersection are:
x = 3/2 + 5t
y = -3/2 - 7t
z = -10t
To calculate the acute angle between the two planes, we can use the dot product of their normal vectors. The dot product is given by cos(theta) = (a · b) / (|a| |b|), where a and b are the normal vectors of the planes. Substituting the values, we have cos(theta) = (2 * 3 + 1 * -4 + -1 * 1) / ((sqrt(2^2 + 1^2 + -1^2))(sqrt(3^2 + -4^2 + 1^2))). Simplifying this equation gives us cos(theta) = 7 / (sqrt(6) * sqrt(26)). Taking the inverse cosine of this value gives us the acute angle between the planes. Using a calculator, we find that the acute angle is approximately 43.77 degrees.